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Is Del (or Nabla, $\nabla$) an operator or a vector ?

\begin{equation*} \nabla\equiv\frac{\partial}{\partial x}\vec{i}+\frac{\partial}{\partial y}\vec{j}+\frac{\partial}{\partial z}\vec{k} \end{equation*}

In some references of vector analysis and electromagnetism, it is considered as an operator (and noted as $\nabla$), and in other ones, it is considered as a vector (and noted as $\vec\nabla$).

valerio
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Sofiane
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  • it is a vector operator. Sometimes the vector part is not noted explicitly, or is dotted (eg \nabla^2 = \nabla \cdot \nabla), but it is definitely a vector operator. – anon01 Oct 10 '16 at 02:07
  • To add to Sean's answer below: It's also a vector in the usual mathematician's sense: a member of the the vector space defined by the set of all linear combinations of the basis vectors $\partial_i$, with the field of scalars being either $\mathbb{R}$ or $\mathbb{C}$ – Selene Routley Oct 10 '16 at 02:59
  • also, notice that $\nabla$ (without the arrow) is usually used to refer to the vector operator (a function from vector to the reals) called the divergence https://en.wikipedia.org/wiki/Divergence –  Oct 10 '16 at 03:19
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    Disagree, @AlbertAspect. In the usage I've seen $\nabla$ is the gradient, $\nabla\cdot$ is the divergence and $\nabla\times$ is the curl. – Sean E. Lake Oct 10 '16 at 03:30

4 Answers4

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First, let's say that $\nabla$ and $\vec \nabla$ are two equivalent notations for the same "object". This notation is used in the representation of three important vector operators: gradient, curl and divergence.

The gradient operator acts on a scalar differentiable function $f(\vec x)$, where $\vec x \in \mathbb R^n$, and returns a vector:

$$\text{grad} \ f(\vec x) = \nabla f(\vec x) \equiv \sum_{i=1}^n \frac{\partial f (\vec x)}{\partial x_i} \vec e_i $$

where $\{\vec e_i \dots\vec e_n\}$ is an orthogonal basis of $\mathbb R^n$.

The divergence operator acts on a vector field $\vec F(\vec x)$, where $\vec x,\vec F \in \mathbb R^n$, and returns a scalar function:

$$\text{div} \ \vec F(\vec x) = \nabla \cdot \vec F(\vec x) \equiv \sum_{i=1}^n \frac{\partial F_i (\vec x)}{\partial x_i} $$

The curl operator acts on a vector field $\vec F(\vec x)$, where $\vec x,\vec F \in \mathbb R^3$, and returns a vector field:

$$\text{curl} \ \vec F(\vec x) = \nabla \times \vec F(\vec x) = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \hat i+\left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \hat j+ \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \hat k$$

where $\hat i, \hat j, \hat k$ are the unit vectors of the three Cartesian axes.

Notice that, unlike the gradient and divergence, the curl operator does not generalize simply in $n$ dimensions. Also, the notation $\nabla \times \vec F$ is only a mnemonic device useful when we work in cartesian coordinates: in other coordinate systems, applying $\nabla \times \vec F$ will hold the wrong result.

We should probably also mention the laplacian operator, which is the divergence of the gradient:

$$\nabla^2 f(\vec x) \equiv \text{div} \ (\text{grad} \ f(\vec x)) = \nabla \cdot (\nabla f(\vec x))$$

So, to sum up, $\nabla$ is just a useful notation that is used in the representation of three different vector operators. It turns out that we can often formally manipulate $\nabla$ as if it was a vector, but it is not a vector in the usual sense: $\nabla$ alone is meaningless.

To see this, just consider one of the fundamental properties of vector spaces: if $v,w$ are elements of the vector space $V$, then $v+w$ is also an element of $V$.

Let's consider the vector space $\mathbb R^n$: what meaning should we give to an expression such as

$$\nabla + \vec x \ ?$$

the answer is: no meaning at all, because $\nabla$ is not a vector.

valerio
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    Good answer, in line with the following from Wikipedia: Strictly speaking, del is not a specific operator, but rather a convenient mathematical notation for those three operators, that makes many equations easier to write and remember. – Peruz Jun 12 '20 at 17:57
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Both. It's an operator that transforms as a covector under rotations. What this means is that if you rotate the coordinate system the gradient in the new coordinate system, $\nabla'$, can be written as:$$\nabla'_i = \sum_{j} R^{-1}_{ij} \nabla_j,$$ where $R^{-1}$ is the inverse of the rotation matrix, $\nabla$ is the gradient in the original coordinate system, and $\nabla'$ is the gradient in the rotated coordinate system.

Sean E. Lake
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    Sorry but can't understand "It's an operator that transforms as a vector under rotations". – Sofiane Oct 10 '16 at 01:54
  • Added elaboration – Sean E. Lake Oct 10 '16 at 02:00
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    Actually it's a covector . . . The matrix should be the inverse of the rotation matrix . . . – Kartik Oct 10 '16 at 02:11
  • How can your rotation matrix-based explanation leads to that $\nabla$ can be both (vector and operator) ? What is the relation between them ? I think that it misses me some basics. – Sofiane Oct 10 '16 at 02:41
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    It's also a vector in the usual mathematician's sense: a member of the the vector space defined by the set of all linear combinations of the basis vectors $\partial_i$, with the field of scalars being either $\mathbb{R}$ or $\mathbb{C}$ – Selene Routley Oct 10 '16 at 02:58
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    It's a linear operator because it takes a function and does something to it. So, if I have a scalar function $\Phi(\mathbf{x})$ then $\nabla \Phi$ is a different, related, function that assigns a vector to every point in space. The linear operator part comes because $\nabla[ a \Phi + b \Psi] = a \nabla \Phi + b \nabla \Psi$ for constant scalars $a$ and $b$, and scalar functions $\Phi$ and $\Psi$. – Sean E. Lake Oct 10 '16 at 03:27
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    @WetSavannaAnimalakaRodVance $\partial_{x_i} \vec e_i$ (I guess you meant to write it with the basis vector) are not vectors, they are differential operators. Also, if a $\nabla$ was an element of the vector space you describe, then an expression like $\nabla \cdot \vec F$ would be meaningless because $\nabla$ and $\vec F$ would belong to different vector spaces. – valerio Oct 11 '16 at 22:24
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    @valerio92 It's standard practice in differential geometry to define the basis vectors of the tangent vector spaces to the manifold as $\operatorname{d}x_i$, and the basis vectors of the covector spaces as $\frac{\partial}{\partial x_i}$. This is done because it makes the directional differential, $\sum_i \operatorname{d}x_i \frac{\partial}{\partial x_i}$, a scalar operator. – Sean E. Lake Oct 12 '16 at 02:33
  • Didn't know about this kind of formalism. I'm reading something, looks like maybe I was wrong and there is a way to interpret $\nabla$ as a (co)vector. – valerio Oct 12 '16 at 08:18
  • @Sofiane Ah, overlooked your comment. There are two different, but related, definitions of a vector. The mathematical definition is that a vector is an element of a vector space, and a vector space is a set with a bunch of properties. Another definition often used in physics is that a (co)vector is an object that transforms under the application of a single (inverse) physical rotation matrix. This is tied in more closely to group theory, and provides a slightly different understanding. – Sean E. Lake Jan 11 '19 at 01:35
  • Can you please elaborate on the definition of vectors in physics or provide a resource explaining it rigorously? I recently had one course on linear algebra so I understand the meaning of vectors in that context but I have trouble understanding in the physics sense. And how are the two definitions in maths and physics related? – Apoorv Potnis May 28 '19 at 10:19
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    @ApoorvPotnis That is a new question, not a comment. Recommended reading: https://en.wikipedia.org/wiki/Linear_form https://www.amazon.com/Mathematics-Classical-Quantum-Physics-Dover/dp/048667164X/ – Sean E. Lake May 28 '19 at 10:34
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    @ApoorvPotnis Also https://physics.stackexchange.com/questions/128026/how-to-understand-the-definition-of-vector-and-tensor , https://physics.stackexchange.com/questions/470998/definition-of-vector – Sean E. Lake May 28 '19 at 11:04
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I hate to play this card, but it depends on the object it acts on (and sometimes who you ask.) Example: many (professors, collegues, etc.) will insist on differentiating between writing $\vec{\nabla}$ and $\nabla$ (consider obliging if your grade/ income depends on it.) In reality, however $\nabla$ is NOT a specific operator, but a convenient mathematical notation. For instance, one may write $\vec{\nabla}\cdot\vec{j}$ or $\nabla\cdot \vec{j}$ and it "should" be obvious from the notation that the meaning of $\nabla$ in this case is a vector operation whether or not the vector symbol is included over it. Another example: one may write $(\vec{v}\cdot\vec\nabla) \vec{j}$ or $\vec{v}\cdot\nabla{\vec {j}}$. In ether case the same quantity is produced. I appreciate the latter notation, however, because it highlights the freedom to act the $\nabla$ upon $\vec{j}$ first (producing a matrix) and then act on $\vec{v}$ to get a vector, or to act the $\vec{v}$ on $\nabla$ first (producing a scalar operator) and then act on $\vec{j}$ producing an identical vector.

IntuitivePhysics
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It's an operator in that it maps vectors from one vector space to vectors in another vector space. The spaces in question here are function spaces.

If you are familiar with Fourier transformations or power series, you know that arbitrary smooth functions can be written as the sum of polynomials or sines and cosines (there are many more sets of functions we can use, but these will suffice to give you the idea). This means that if I have some smooth function $f(x)$, I can write it as

$$ f(x) = a_1 sin(\frac{\pi x}{L}) + a_2 sin(\frac{2 \pi x}{L}) + ... + b_1 cos(\frac{\pi x}{L})+ ... $$

It should not be a huge leap to see then that $f(x)$ can be represented as a vector with entries

$$ \langle a_1, a_2,..., b_1, ...\rangle. $$

The derivative of $f(x)$ is then another vector in this space since it also has a Fourier transformation. This is the sense in which $\nabla$ is an operator. It takes vectors in a function space to vectors in a function space.

Now, given a function of several variables, we could take derivatives of that fucntion along arbitrary combinations of those variables. For instance, given $f(x,y)$ we can take $\frac{\partial}{\partial y}$ or $\frac{\partial}{\partial x}$, or even do something crazy like define $z=\frac{x}{sin(y)}$ and try take $\frac{d}{d z}$. We won't be doing the last thing...

Instead we focus only on "directional derivatives". First recall that linear combinations of $x$ and $y$ define an arrow in $\mathbb{R}^2$, that is the arrow from the origin to $z=a \hat{x} + b \hat{y}$.

One question we might want to ask is "what is the space of directional derivatives of $f(x,y)$." It is similar to the space of $z$s, but since we're talking about derivatives we'll write the vectors in $\mathbb{R}^2$ as $ a \frac{\partial}{\partial x} + b \frac{\partial}{\partial y}$. This is the sense in which $\nabla$ is a vector, because the space of directional derivatives of functions of $N$ variables (equally functions on $\mathbb{R}^N$) is $\mathbb{R}^N$.

So as others have pointed out, it is both an operator and a vector. It is a vector in the space of directional derivatives on functions of $N=3$ variables, and it is an operator on functions of $N=3$ variables.

CasualScience
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